## digitalmars.D.learn - floating point divide

- "Damian" <damianday hotmail.co.uk> Oct 11 2012
- "Aziz K." <aziz.koeksal gmail.com> Oct 11 2012
- "bearophile" <bearophileHUGS lycos.com> Oct 11 2012
- "Damian" <damianday hotmail.co.uk> Oct 11 2012

I come from a pascal background and we could use: div integral division operator / floating point division operator So my question is, how does D force floating point division on integrals? At the moment i do this, but i was hoping for an easier way: int n1 = 10, n2 = 2; float f = cast(float)(cast(float)(n1 / n2));

Oct 11 2012

int n1 = 10, n2 = 2; float f = (n1+0.0f)/n2; Casting n1 to float would also work, but I hope the compiler is smart enough to optimize away the plus expression.

Oct 11 2012

Damian:I come from a pascal background and we could use: div integral division operator / floating point division operator

Two operators for the two different operations is a design better than C, that is bug-prone.So my question is, how does D force floating point division on integrals? At the moment i do this, but i was hoping for an easier way: int n1 = 10, n2 = 2; float f = cast(float)(cast(float)(n1 / n2));

That's not good, it performs an integer division, followed by two float casts. Note: float is useful only if you have many of them, or if you pass/return pairs of them. A single float is not so useful. A solution: int n1 = 10, n2 = 2; const f = n1 / cast(double)n2; Using type inference is useful, as it doesn't hide an integer result if your code is wrong. Bye, bearophile

Oct 11 2012

On Thursday, 11 October 2012 at 15:21:01 UTC, bearophile wrote:Damian:I come from a pascal background and we could use: div integral division operator / floating point division operator

Two operators for the two different operations is a design better than C, that is bug-prone.So my question is, how does D force floating point division on integrals? At the moment i do this, but i was hoping for an easier way: int n1 = 10, n2 = 2; float f = cast(float)(cast(float)(n1 / n2));

That's not good, it performs an integer division, followed by two float casts. Note: float is useful only if you have many of them, or if you pass/return pairs of them. A single float is not so useful. A solution: int n1 = 10, n2 = 2; const f = n1 / cast(double)n2; Using type inference is useful, as it doesn't hide an integer result if your code is wrong. Bye, bearophile

Ah i see, thankyou for the explanation.

Oct 11 2012